nd Midterm Determine syndrome vectors for single bit errors for the code

Nd midterm determine syndrome vectors for single bit

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88.02.12.2011 2ndMidterm Determine syndrome vectors for single bit errors for the code described by 3 ( ) 1 g p p p . Solution Single bit errors can be simulated by 6 5 1 , , , ,1 p p p as “0000000” is one of the codewords. Dividing each of these polynomials by ( ) g p and taking the remainder we find syndrome polynomials, hence vectors. Other syndrome vectors are calculated similarly and we get S = {“101”,”111”, “110”,”011”,”100”,”010”,”001”} 89. 02.12.2011 2 nd Midterm Determine if (7,4) Hamming code is linear. Parity equations are given as 5 1 2 44 c c c c , 6 1 3 c c c c , 7 2 3 4 c c c c Solution First four bits are straight binary information words (systematic). Their modulo-sums make up other valid information words. What we need to check is whether or not their parity bits can be obtained by summing the parity bits for the information words that are summed. That is, let C x =x 1 x 2 x 3 x 4 p x1 p x2 p x3 and C y =y 1 y 2 y 3 y 4 p y1 p y2 p y3 be given for information words X =x 1 x 2 x 3 x 4 and Y =y 1 y 2 y 3 y 4 . We know that Z=X+Y since they are just binary codes. The question is whether p zi = p xi + p yi holds or not. Writing Z=X+Y for parity bit p z1 and using 5 1 2 4 c c c c we get 1 2 4 1 2 4 1 2 4 z z z x x x y y y which always holds only when 1 1 1 z x y , 2 2 2 z x y and 4 4 4 z x y . Similarly for p z2 and 1 3 4 1 3 4 1 3 4 z z z x x x y y y we need to have 1 1 1 z x y , 3 3 3 z x y and 4 4 4 z x y . Finally for p z3 and 2 3 4 2 3 4 2 3 4 z z z x x x y y y , the equations 2 2 2 z x y , 3 3 3 z x y and 4 4 4 z x y must hold. QED. 6 p 3 1 p p 6 4 3 p p p 3 1 p p 4 3 p p 4 2 p p p 3 2 p p p 3 p p p 2 1 p “101”
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151227621 DIGITAL COMMUNICATIONS 53 90. 02.12.2011 2 nd Midterm Draw state transition graph for the rate 1/2 convolutional encoder given with 2 1 2 3 i i i i i O x x x x , 2 1 1 3 i i i i O x x x . Mark output values on the graph. Solution 91. 02.12.2011 2 nd Midterm While processing received sequence on trellis diagram we keep only one of the incoming branches at every step even when the removed branch metrics are lower than that of the incoming branches at other states. Why? Solution Independent from the metrics calculated for the departing branches, lowest one that passes an individual node will be the one that uses the branch with smallest metric at that node. Therefore, there is no point keeping both numbers. In the example given, it does not matter whichever of the branches to C and D adds smaller cost, they will both include the branch that passes from B. None of the branches that depart from node X will use the dashed branch (from A), so dashed branch can safely be deleted. 000 001 010 011 100 101 110 111 000 001 010 011 100 101 110 111 00 11 10 01 11 01 10 11 00 01 10 00 11 10 01 00 O 2i x i x i-1 x i-2 x i-3 O 2i+1 in-stream 3 2 0 2 A B C D X
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151227621 DIGITAL COMMUNICATIONS 54 92. 12.01.2012 Final Exam In 1-bit error correcting block codes (Hamming codes, for example), the minimum Hamming distance or minimum weight of the code is 3.
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  • Mr. Bhullar
  • Hamming Code, Error detection and correction, Parity bit

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